Question

Consider the reaction sequence Dimethyl ketone \( \xrightarrow{\text{(i)CH}_3\text{MgCl (ii)H}_2\text{O}} \) X \( \xrightarrow{\text{(i)Na (ii)CH}_3\text{Br}} \) Y How many sp\(^3\) carbons are present in Y?

• A. 5
• B. 4
• C. 3
• D. 6

Hint:

Reaction sequence steps: 1. Ketone + Grignard reagent (\(R'MgX\)) followed by hydrolysis (\(H_2O\)): Forms a tertiary alcohol if the ketone is not formaldehyde or other aldehyde. \( R_1COR_2 + R'MgX \rightarrow R_1R_2R'C-OMgX \xrightarrow{H_2O} R_1R_2R'C-OH \). Acetone (\(CH_3COCH_3\)) + \(CH_3MgCl\) \( \rightarrow \) tert-Butyl alcohol (\((CH_3)_3COH\)). 2. Alcohol + Na: Forms sodium alkoxide (\(RO^-Na^+\)). 3. Sodium alkoxide + Alkyl halide (Williamson Ether Synthesis): \(RO^-Na^+ + R'X \rightarrow ROR' + NaX\). Best with primary alkyl halides (R'X) to favor S\(_N\)2 over elimination. Hybridization: - Carbon atom with 4 single bonds: sp\(^3\). - Carbon atom with 1 double bond and 2 single bonds: sp\(^2\). - Carbon atom with 1 triple bond and 1 single bond, or 2 double bonds: sp.

Answer 1

The Correct Option is A

Dimethyl ketone is acetone (\( \text{CH}_3\text{COCH}_3 \)).
Step 1: Dimethyl ketone \( \xrightarrow{\text{(i)CH}_3\text{MgCl (ii)H}_2\text{O}} \) X This is the reaction of a ketone with a Grignard reagent (CH\(_3\)MgCl), followed by hydrolysis.
Acetone (\( (\text{CH}_3)_2\text{C=O} \)) reacts with CH\(_3\)MgCl: The CH\(_3^-\) from Grignard attacks the carbonyl carbon.
\[ (\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{MgCl} \rightarrow (\text{CH}_3)_2\text{C(OMgCl)CH}_3 \] Hydrolysis (\( \text{H}_2\text{O} \)) protonates the alkoxide: \[ (\text{CH}_3)_2\text{C(OMgCl)CH}_3 \xrightarrow{\text{H}_2\text{O}} (\text{CH}_3)_2\text{C(OH)CH}_3 + \text{Mg(OH)Cl} \] Product X is tert-Butyl alcohol (2-methylpropan-2-ol): \( (\text{CH}_3)_3\text{C-OH} \).
Structure of X: Central carbon bonded to three CH\(_3\) groups and one -OH group.
All four carbons are sp\(^3\) hybridized.
Step 2: X \( \xrightarrow{\text{(i)Na (ii)CH}_3\text{Br}} \) Y X is tert-Butyl alcohol (\( (\text{CH}_3)_3\text{C-OH} \)).
Reaction (i) with Na: Alcohols react with sodium metal to form sodium alkoxides and hydrogen gas.
\[ (\text{CH}_3)_3\text{C-OH} + \text{Na} \rightarrow (\text{CH}_3)_3\text{C-O}^-\text{Na}^+ + \frac{1}{2}\text{H}_2\text{(g)} \] Intermediate is sodium tert-butoxide.
Reaction (ii) with CH\(_3\)Br: Sodium tert-butoxide reacts with methyl bromide (CH\(_3\)Br).
This is a Williamson ether synthesis.
The tert-butoxide ion (\( (\text{CH}_3)_3\text{C-O}^- \)) is a strong base but also a nucleophile (though hindered).
With a primary alkyl halide like CH\(_3\)Br, S\(_N\)2 reaction can occur to form an ether.
\[ (\text{CH}_3)_3\text{C-O}^-\text{Na}^+ + \text{CH}_3\text{Br} \rightarrow (\text{CH}_3)_3\text{C-O-CH}_3 + \text{NaBr} \] Product Y is methyl tert-butyl ether (MTBE).
Structure of Y: \( (\text{CH}_3)_3\text{C-O-CH}_3 \).
This is a central carbon (from tert-butyl group) bonded to three CH\(_3\) groups and one -O-CH\(_3\) group.
The carbons are: - Three methyl carbons of the tert-butyl group: Each is sp\(^3\).
(3 sp\(^3\) carbons) - The central quaternary carbon of the tert-butyl group: It is bonded to 3 other carbons and 1 oxygen.
It is sp\(^3\).
(1 sp\(^3\) carbon) - The methyl carbon from -O-CH\(_3\): It is bonded to 3 hydrogens and 1 oxygen.
It is sp\(^3\).
(1 sp\(^3\) carbon) Total number of sp\(^3\) hybridized carbons in Y = \(3+1+1 = 5\).
This matches option (1).