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Statement 1: Reduction of 3-phenyl prop-...

Statement 1: Reduction of `3`-phenyl prop-2-en-1-al with `LAH` gives 3-phenyl propan-1-ol.
Statement 2: Both the double bond and the aldehyde group of `alpha, beta`-unsaturated aldehydes are reduced by `LAH`.

A

a.Statement 1 and Statement 2 are true and Statement 2 is the correct explanation of Statement 1.

B

b.Statement 1 and Statement 2 are true and Statement 2 is not the correct explanation of Statement 1.

C

c.Statement 1 is true and statement 2 is false.

D

d.Statement 1 is false and statement 2 is false

Text Solution

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To solve the problem, we will analyze both statements regarding the reduction of 3-phenyl prop-2-en-1-al with lithium aluminum hydride (LAH) and determine their validity step by step. ### Step 1: Understand the Compound The compound in question is 3-phenyl prop-2-en-1-al. It has the following structure: - The aldehyde group (-CHO) is located at the first carbon. - The double bond (C=C) is between the second and third carbons. - A phenyl group (C6H5) is attached to the third carbon. ### Step 2: Identify the Reducing Agent Lithium aluminum hydride (LAH) is a strong reducing agent. It is capable of reducing: - Aldehydes and ketones to their corresponding alcohols. - Alkenes to alkanes. ### Step 3: Analyze Statement 1 Statement 1 claims that the reduction of 3-phenyl prop-2-en-1-al with LAH gives 3-phenyl propan-1-ol. - When LAH is applied to the compound, it will reduce the aldehyde group to an alcohol, converting it from -CHO to -CH2OH. - Additionally, the double bond (C=C) between the second and third carbons will be reduced to a single bond (C-C). The resulting structure will be: - The aldehyde becomes a primary alcohol (-CH2OH). - The double bond is saturated, resulting in a fully saturated carbon chain. Thus, the final product is indeed 3-phenyl propan-1-ol. ### Step 4: Analyze Statement 2 Statement 2 states that both the double bond and the aldehyde group of alpha, beta-unsaturated aldehydes are reduced by LAH. - In the case of 3-phenyl prop-2-en-1-al, the compound is an alpha, beta-unsaturated aldehyde. - LAH will reduce both the aldehyde group and the double bond, as previously discussed. ### Conclusion Both statements are correct: - Statement 1 is true because the reduction yields 3-phenyl propan-1-ol. - Statement 2 is true because LAH reduces both the double bond and the aldehyde group in alpha, beta-unsaturated aldehydes. ### Final Answer Both statements are correct, and Statement 2 is the correct explanation for Statement 1. Therefore, the answer is option A. ---
To solve the problem, we will analyze both statements regarding the reduction of 3-phenyl prop-2-en-1-al with lithium aluminum hydride (LAH) and determine their validity step by step. ### Step 1: Understand the Compound The compound in question is 3-phenyl prop-2-en-1-al. It has the following structure: - The aldehyde group (-CHO) is located at the first carbon. - The double bond (C=C) is between the second and third carbons. - A phenyl group (C6H5) is attached to the third carbon. ...
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