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Chapter 6: Problem 62

A compound (X) of formula \(\mathrm{C}_{3} \mathrm{H}_{0} \mathrm{~N}\) was treated with chloroform and alcoholic KOH. The organic product formed was reduced to give a secondary amine. Identify the true statement regarding \((\mathrm{X})\) (a) It exists in only one form which is optically inactive. (b) It exists in two isomeric forms and both the forms are optically inactive. (c) It exists in two forms and only one form shows optical activity. (d) It exists in two forms and both forms exhibit optical isomerism.

Short Answer

Expert verified
Based on the given molecular formula (C3H3N) and reactions with chloroform and alcoholic KOH, the compound (X) is propenenitrile. It can exist in two forms: E-isomer, which is symmetric and does not show optical activity, and Z-isomer, which is asymmetric and contains a chiral center, thus showing optical activity. Therefore, the correct statement regarding (X) is: (c) It exists in two forms, and only one form shows optical activity.

Step by step solution

01

Deduce the Structure of the Compound(X)

First, we need to deduce the structure of Compound (X). Using the molecular formula \(\mathrm{C}_{3} \mathrm{~H}_{0} \mathrm{~N}\), we can conclude that the structure contains one nitrogen atom with two carbon atoms bonded and one carbon atom as a triple bond to the nitrogen (a nitrile group) since there are no hydrogens. So, the possible structure of Compound (X) is a propenenitrile.
02

Reaction with Chloroform and Alcoholic KOH

Compound (X) reacts with chloroform and alcoholic KOH to give an organic product. According to this reaction's mechanism, it is a nucleophilic aliphatic substitution reaction (SN2) since there's a nucleophile (alcoholic KOH) involved. The alcohol from KOH, as the nucleophile, will attack the atom with the highest partial positive charge (carbon bonded to nitrogen in case of nitrile).
03

Deduction of the Organic Product's Structure and Reduction

After the completion of the nucleophilic aliphatic substitution reaction in step 2, the resulting structure would be that of a Haloalkane, followed by the removal of hydrogen halide (HCl). The reduction of the organic product ultimately yields a secondary amine.
04

Optical Isomerism in Compound(X)

Now, we need to determine optical isomerism in the possible forms of Compound(X). The compound (X) in question contains a carbon-carbon double bond (C=C) that can exist as E or Z isomers (cis-trans isomers). The different forms of (X) are as follows: 1. The E-isomer of propenenitrile: The molecule is symmetric and does not contain a chiral center; thus, it is optically inactive. 2. The Z-isomer of propenenitrile: The molecule is asymmetric and contains one chiral center, resulting in optical isomerism.
05

Identifying the Correct Statement

Using the above analysis, it is clear that compound (X) can exist in two forms, and only one form shows optical activity. Therefore, the correct statement regarding (X) is: (c) It exists in two forms, and only one form shows optical activity.

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