Very important questions for engineering chemistry in b.tech 1st year.
So he guys! Today this post about that is very important questions for chemistry in b.tech first year. Most imp questions for semester exam in chemistry subject.
Unit 5 :- organic reaction and synthesis of drugs molecular
Que.1:- Explain SN1 and SN2 substitution reaction.
Ans:-
SN1 and SN2 are two common types of nucleophilic substitution reactions that occur in organic chemistry.
SN1 (Substitution Nucleophilic Unimolecular) reaction is a two-step process where the leaving group departs from the substrate to form a carbocation intermediate, which is then attacked by a nucleophile to form the product. This reaction is characterized by the involvement of a carbocation intermediate, which is formed in the slow, rate-determining step. Since the intermediate is highly unstable and can rearrange to form more stable products, SN1 reactions often exhibit racemization and/or carbocation rearrangement. These reactions usually occur in polar protic solvents like water or alcohol and with substrates that have weak nucleophiles and strong leaving groups.
SN2 (Substitution Nucleophilic Bimolecular) reaction is a one-step process where the nucleophile attacks the substrate at the same time as the leaving group departs. The reaction proceeds through a transition state where the nucleophile and the leaving group are both present on the substrate. This reaction is characterized by a concerted mechanism in which the bond breaking and bond forming occur simultaneously. SN2 reactions usually occur in polar aprotic solvents like DMF or DMSO and with substrates that have strong nucleophiles and weak leaving groups.
Overall, the key difference between SN1 and SN2 reactions lies in the number of steps involved in the reaction and the nature of the substrate, nucleophile, and leaving group. SN1 reactions proceed via a carbocation intermediate and are favored with weak nucleophiles and strong leaving groups, while SN2 reactions proceed via a concerted mechanism and are favored with strong nucleophiles and weak leaving groups.
Que.2:- What is electrophilic addition reaction? Explain markownikoff rule with machanism.
Ans:-
Electrophilic addition reaction is a type of organic reaction in which an electrophile adds to a molecule or substrate, usually a double or triple bond, to form a new single bond. This reaction is commonly observed in alkenes and alkynes, which have pi bonds that are susceptible to attack by electrophiles.
The Markovnikov rule, named after Russian chemist Vladimir Markovnikov, states that in an electrophilic addition reaction of a protic acid (HX) to an asymmetrically substituted alkene or alkyne, the hydrogen atom of the acid will add to the carbon atom that has the greater number of hydrogen atoms already attached to it, while the X (halogen) atom will add to the other carbon atom.
The mechanism of the Markovnikov rule in the addition of HX to an unsymmetrical alkene can be explained as follows:
1. The electrophilic H+ ion in HX is attracted to the pi bond of the alkene, forming a carbocation intermediate.
2. The carbocation intermediate can be stabilized by resonance, where the positive charge is delocalized over the double bond, resulting in a more stable intermediate.
3. The nucleophilic X- ion attacks the carbocation intermediate, forming a new single bond and completing the addition reaction.
4. In accordance with the Markovnikov rule, the hydrogen atom of the HX molecule will add to the carbon atom that has the most hydrogen atoms already attached to it, resulting in the formation of the more stable carbocation intermediate. Therefore, the X- ion will add to the other carbon atom.
For example, in the addition of HBr to propene, the H+ ion of HBr will add to the carbon atom of the double bond that has two hydrogen atoms attached to it, while the Br- ion will add to the other carbon atom, as shown below:
CH3CH=CH2 + HBr → CH3CH(Br)CH3
This reaction follows the Markovnikov rule, where the hydrogen atom of HBr adds to the less substituted carbon atom, while the bromine atom adds to the more substituted carbon atom, resulting in the formation of 2-bromopropane.
Que.3:- Explain paraoxide effect with machanism.
Ans:-
The Paracetaldehyde (also known as paraoxide) effect is a phenomenon that occurs during the oxidation of primary alcohols to aldehydes. This effect refers to the observation that the reaction rate is increased when a small amount of aldehyde is added to the reaction mixture. The effect is named after the compound paracetaldehyde, which is often used to facilitate the oxidation of primary alcohols to aldehydes.
The mechanism of the Paracetaldehyde effect can be explained as follows:
1. The oxidation of primary alcohols to aldehydes is a two-step process, involving the formation of an aldehyde intermediate before the final oxidation to the carboxylic acid.
2. During the first step, the primary alcohol is oxidized to an aldehyde by a strong oxidizing agent such as pyridinium chlorochromate (PCC).
3. However, the oxidation of the aldehyde intermediate to the carboxylic acid is typically much slower than the initial oxidation of the primary alcohol.
4. The addition of a small amount of an aldehyde such as paracetaldehyde to the reaction mixture increases the rate of the oxidation of the aldehyde intermediate to the carboxylic acid.
5. This is because the aldehyde acts as a catalyst, facilitating the oxidation of the aldehyde intermediate by the oxidizing agent.
6. The aldehyde catalyzes the oxidation reaction by forming an intermediate complex with the oxidizing agent, which then transfers an oxygen atom to the aldehyde intermediate to form the carboxylic acid product.
Overall, the Paracetaldehyde effect is an example of catalysis by a small molecule, where the aldehyde acts as a catalyst to increase the rate of the oxidation reaction by facilitating the transfer of an oxygen atom from the oxidizing agent to the aldehyde intermediate. This effect is particularly useful in organic synthesis as it allows for the efficient and selective conversion of primary alcohols to aldehydes.
Que.4:- What is elimination reaction? Explain unimolecular (E1) and bimolecular (E2) elimination reaction with machanism.
Ans:-
Elimination reactions are a type of organic reaction in which a molecule loses a small molecule such as water or hydrogen halide to form a double bond or triple bond. There are two main types of elimination reactions: unimolecular (E1) and bimolecular (E2) elimination reactions.
Unimolecular (E1) elimination reaction:
In E1 elimination, the reaction proceeds in two steps, the first step is the slowest and is the formation of a carbocation intermediate, while the second step is the fast loss of a leaving group (such as water or halide) from the carbocation intermediate, leading to the formation of an alkene or alkyne product. The mechanism for E1 elimination is as follows:
The leaving group (X) dissociates from the substrate, forming a carbocation intermediate.
The carbocation intermediate is then stabilized by resonance or neighboring groups that can donate electron density, leading to a more stable carbocation intermediate.
A proton is then abstracted by a base to form a double bond or a triple bond and regenerate the base for the next reaction cycle.
The rate-determining step for this reaction is the formation of the carbocation intermediate, making the reaction dependent on the stability of the carbocation intermediate. The reaction proceeds faster when the intermediate is more stable.
Bimolecular (E2) elimination reaction:
In E2 elimination, the reaction occurs in a single concerted step where a proton is removed from the substrate by a base and the leaving group departs simultaneously. The mechanism for E2 elimination is as follows:
A strong base abstracts a proton from the β-carbon (adjacent to the leaving group) of the substrate, creating a transition state in which the C-H bond is partially broken and the C-LG bond is partially formed.
The leaving group (LG) departs from the substrate, forming a double bond or a triple bond in the product.
The rate of the E2 elimination reaction is dependent on both the concentration of the substrate and the base, as well as the steric hindrance around the β-carbon, with less hindered substrates reacting faster.
In summary, unimolecular (E1) and bimolecular (E2) elimination reactions are important organic reactions that involve the loss of a small molecule to form a double bond or a triple bond. The two types of reactions differ in the number of steps and the rate-determining step involved .
Que.5:- Explain Wolff Kishner reaction with machanism.
Ans:-
The Wolff-Kishner reaction is a chemical reaction used to convert carbonyl compounds, such as aldehydes and ketones, into alkanes. The reaction involves the use of hydrazine (N2H4) and a strong base, typically potassium hydroxide (KOH), to generate the desired product.
The mechanism of the reaction is as follows:
Step 1: Formation of the hydrazone
The carbonyl compound is first reacted with hydrazine to form a hydrazone. This reaction is catalyzed by a weak acid such as acetic acid (CH3COOH).
R1-C=O + N2H4 → R1-C=N-NH2 + H2O
Step 2: Formation of the alkane
The hydrazone is then treated with a strong base, typically potassium hydroxide (KOH), to form the desired alkane. The strong base deprotonates the hydrazone, generating an anionic intermediate.
R1-C=N-NH2 + 2 KOH → R1-CH3 + N2 + 2 H2O + 2 K+
This intermediate is then treated with the same hydrazine molecule that was used in the first step, to regenerate the hydrazone and release the alkane product.
R1-C=N-NH2 + N2H4 → R1-CH3 + N2 + 2 NH3
Overall reaction:
R1-C=O + 2 N2H4 + KOH → R1-CH3 + N2 + 3 H2O + K+
In summary, the Wolff-Kishner reaction involves the conversion of a carbonyl compound to an alkane through the formation of a hydrazone intermediate, which is then treated with a strong base to release the desired product.
Que.6:- Reimer tiemann reaction with machanism.
Ans:-
The Reimer-Tiemann reaction is a reaction that involves the conversion of an aromatic compound (usually phenol or aniline) into a salicylaldehyde or an ortho-substituted phenol, respectively. The reaction is named after the chemists Karl Reimer and Ferdinand Tiemann, who first described it in 1876.
The mechanism of the Reimer-Tiemann reaction involves the following steps:
• Formation of the Reimer-Tiemann intermediate: The reaction is initiated by the addition of a strong base, typically sodium hydroxide, to a mixture of phenol and chloroform. The hydroxide ion (OH-) deprotonates the phenol to form the phenoxide ion (PhO-). The phenoxide ion then reacts with chloroform to form the Reimer-Tiemann intermediate, which is a carbonyl-containing species that has a negative charge on the carbon atom.
• Rearrangement of the Reimer-Tiemann intermediate: The intermediate undergoes a rearrangement to form the salicylaldehyde or ortho-substituted phenol. In the case of phenol, the intermediate rearranges to form salicylaldehyde, whereas in the case of aniline, the intermediate rearranges to form an ortho-substituted phenol.
Overall, the Reimer-Tiemann reaction can be summarized as follows:
PhOH + CHCl3 + NaOH → PhO-CHCl2 → PhOCHO + HCl
The reaction can be catalyzed by Lewis acids such as AlCl3, which can facilitate the formation of the intermediate and accelerate the reaction. The Reimer-Tiemann reaction is an important tool for the synthesis of salicylaldehydes and ortho-substituted phenols, which have various applications in the fields of medicine, agriculture, and materials science.
Que.7:- Cannizaro reaction with machanism.
Ans:-
The Cannizzaro reaction is a chemical reaction that involves the oxidation-reduction of aldehydes. It is named after the Italian chemist Stanislao Cannizzaro, who first described the reaction in 1853.
The mechanism of the Cannizzaro reaction involves the following steps:
Deprotonation: The reaction is initiated by the deprotonation of an aldehyde molecule by a strong base such as sodium hydroxide. The deprotonated aldehyde forms a carbanion intermediate.
Nucleophilic attack: Another molecule of the aldehyde then acts as a nucleophile, attacking the carbanion intermediate and forming an unstable intermediate known as a hemiacetal.
Proton transfer: The hemiacetal intermediate then undergoes a proton transfer reaction, in which a proton is transferred from the alcohol group of the hemiacetal to the carbanion.
Rearrangement: The resulting carbanion intermediate then undergoes a rearrangement to form an alcohol and a carboxylic acid. The alcohol formed is the reduced product, while the carboxylic acid is the oxidized product.
Overall, the Cannizzaro reaction can be summarized as follows:
2 RCHO + NaOH → RCOOH + RCH2OH + Na+
The Cannizzaro reaction is used for the synthesis of primary alcohols and carboxylic acids from aldehydes. The reaction is often used in the production of fragrances and flavors in the food industry. The reaction can also be used for the determination of the structure of aldehydes by identifying the products formed.
Que.8:- Wurtz reaction with machanism.
Ans:-
The Wurtz reaction is a chemical reaction that involves the coupling of two alkyl halides to form a higher molecular weight alkane. It is named after the German chemist Charles Adolphe Wurtz, who first described the reaction in 1855.
The mechanism of the Wurtz reaction involves the following steps:
Formation of an alkyl radical: The reaction is initiated by the reduction of an alkyl halide with metallic sodium. The sodium removes the halogen atom from the alkyl halide, generating an alkyl radical.
Formation of a sodium alkyl: The alkyl radical then reacts with another molecule of the alkyl halide to form a sodium alkyl intermediate. The sodium alkyl intermediate is stabilized by coordination of the sodium ion to the alkyl group.
Coupling: The sodium alkyl intermediate then undergoes a coupling reaction, in which two sodium alkyl intermediates react to form a higher molecular weight alkane.
Regeneration of sodium: Finally, the reaction is completed by the regeneration of sodium halide from the reaction of sodium and the halogen atom released in step 1.
Overall, the Wurtz reaction can be summarized as follows:
2 R-X + 2 Na → R-R + 2 NaX
where R represents an alkyl group, X represents a halogen atom, and NaX represents the corresponding sodium halide.
The Wurtz reaction is a useful method for the synthesis of symmetric alkanes, but it is limited by the need for two identical alkyl halides to couple. The reaction is often used in the synthesis of hydrocarbons, as well as in the preparation of starting materials for organic synthesis.
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This is a very important questions for semester exam in b.tech first year.
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