The Question is from neet-chemistry organic-chemistry-basics , and it was created by the author Mr. S M HUSSAIN

Question

An organic compound with molecular formula $C_7H_{14}O$ exhibits optical isomerism but not geometrical isomerism. It also gives a positive iodoform test. Which of the following could be the compound?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D
## CONCEPT FRAMEWORK
 **Primary Concept**: Stereochemistry, Iodoform test
 **Related Topics**: Optical isomerism, Geometrical isomerism, Ketones, Aldehydes
 **Prerequisites**: Understanding of functional groups, chirality, and the iodoform test.
## SOLUTION PATHWAY
**Given Information**:
  Molecular formula: $C_7H_{14}O$
  Exhibits optical isomerism (presence of chiral center)
  Does not exhibit geometrical isomerism (no double bonds)
  Positive iodoform test (presence of a methyl ketone group, $CH_3CO-$)
**Method & Steps**:
 1. Analyze the molecular formula $C_7H_{14}O$. The degree of unsaturation is 1, indicating one double bond or a ring. Since geometrical isomerism is absent, it must be a ring or a ketone.
  Reasoning: The formula suggests a ketone or a cyclic alcohol.  The absence of geometrical isomerism rules out alkenes.
  Formula/Rule: Degree of unsaturation = $\frac{2C + 2 + N - X - H}{2}$, where C is carbon, N is nitrogen, X is halogen, and H is hydrogen.
 2. Consider the positive iodoform test. This confirms the presence of a methyl ketone ($CH_3CO-$) group.
  Working: The iodoform test is specific to methyl ketones and methyl secondary alcohols that can be oxidized to methyl ketones.
  Key Point:  Only methyl ketones give a positive iodoform test.
 3. Examine the options to identify a methyl ketone with seven carbons, showing optical isomerism but no geometrical isomerism. 4-methylhexan-3-one fits this description.
  Result: 4-methylhexan-3-one
  Verification: It has a methyl ketone group, a chiral carbon (carbon 3), and no double bonds.
 ## OPTION ANALYSIS
[A] Heptan-2-one :x:  It's a methyl ketone, but it lacks a chiral center.
[B] 3-methylhex-3-en-2-one :x: It has a methyl ketone and a double bond, leading to geometrical isomerism.
[C] 2,4-dimethylpentan-3-one :x: It's a ketone but lacks a chiral center.
[D] 4-methylhexan-3-one ✓: It's a methyl ketone with a chiral center (carbon 3) and no double bonds, fulfilling all criteria.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the iodoform test's specificity:  Many ketones don't give a positive test.
  2.Misinterpreting the degree of unsaturation: This can lead to incorrect functional group identification.
 **Quick Tips**:
  1.Draw out the structures of the options to easily visualize the presence/absence of chiral centers and double bonds.
  2.Remember the key characteristics of the iodoform test.

Mr. S M HUSSAIN · Answer

Heptan-2-one

Mr. S M HUSSAIN · Answer

3-methylhex-3-en-2-one

Mr. S M HUSSAIN · Answer

2,4-dimethylpentan-3-one

Mr. S M HUSSAIN · Answer

The correct Option is "4-methylhexan-3-one"

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