To solve this problem, we need to find all isomeric ketones (including stereoisomers) with molecular weight 100 that, when reduced with NaBH₄, produce racemic products. NaBH₄ reduces ketones to secondary alcohols. A racemic mixture is formed if the reduction creates a new chiral center, and the starting ketone is achiral (so the reaction produces both enantiomers equally). If the ketone is already chiral, reduction might give diastereomers, not racemates. But note: the problem asks for racemic products, meaning the product itself is racemic.

Step 1: Determine the molecular formula for ketones with MW=100.
General formula for ketone is C_nH_2nO. So, molecular weight = 12n + 2n + 16 = 14n + 16 = 100.
$14n+16=100\Rightarrow 14n=84\Rightarrow n=6$
So, the molecular formula is C₆H₁₂O.

Step 2: Find all possible ketone isomers with C₆H₁₂O.
We consider all structural isomers (chain, position) and stereoisomers (geometrical, optical).

A. Chain isomers:
(i) Hexanone: Two positional isomers - hexan-2-one and hexan-3-one.
(ii) Branched chain:

• 2-methylpentan-3-one
• 2-methylpentan-2-one? But this is not possible as it would be a tertiary alcohol? Wait, ketones have C=O, so 2-methylpentan-2-one is actually a ketone? Actually, 2-methylpentan-2-one would have formula C₆H₁₂O? Let's check: Carbonyl carbon is C2, with methyl and pentyl? Actually, it is valid: CH₃-CO-CH(CH₃)-CH₂-CH₃? But that is 3-methylbutanone? Wait, systematic naming: The longest chain is 5 carbons? Actually, 2-methylpentan-2-one is not a valid ketone because the carbonyl carbon would be C2, but it would be attached to two methyl groups and a propyl? That is actually 3-methylbutan-2-one? Let's list properly:

Actually, for ketones with 6 carbons, the possible skeletons are:

• Straight chain: hexan-2-one, hexan-3-one
• Branched:
  • 3-methylpentan-2-one
  • 4-methylpentan-2-one
  • 3-methylpentan-3-one? But 3-methylpentan-3-one would be CH₃CH₂-C(CH₃)(O)-CH₂CH₃? Actually, it is valid.
  • 2-methylpentan-3-one
  • 4-methylpentan-3-one? Same as 2-methylpentan-3-one? Actually, careful: 4-methylpentan-2-one and 4-methylpentan-3-one are different.
  Actually, to avoid missing, we list all constitutional isomers:

There are 6 constitutional isomers for ketones with C₆H₁₂O:

1. Hexan-2-one
2. Hexan-3-one
3. 3-methylpentan-2-one
4. 4-methylpentan-2-one
5. 3-methylpentan-3-one
6. 2-methylpentan-3-one

Note: 2,2-dimethylbutan-3-one? That would be C₆H₁₂O? Actually, CH₃-CO-C(CH₃)₃ is 3,3-dimethylbutan-2-one, which has formula C₆H₁₂O. So we missed one: 3,3-dimethylbutan-2-one. So total 7 constitutional isomers.

So, 7 constitutional isomers:

1. Hexan-2-one
2. Hexan-3-one
3. 3-methylpentan-2-one
4. 4-methylpentan-2-one
5. 3-methylpentan-3-one
6. 2-methylpentan-3-one
7. 3,3-dimethylbutan-2-one

Step 3: Consider stereoisomers.
We need to check for geometrical isomers (if alkene-like) and chiral centers.

• Hexan-2-one: No stereoisomers (no chiral center, no geometrical isomerism).
• Hexan-3-one: No stereoisomers.
• 3-methylpentan-2-one: The carbon adjacent to carbonyl? Actually, check for chiral center: The carbon at position 3 is chiral? In 3-methylpentan-2-one, the structure is CH₃-CO-CH(CH₃)-CH₂CH₃. So, carbon 3 is chiral (attached to H, CH₃, COCH₃, CH₂CH₃). So, it has a pair of enantiomers.
• 4-methylpentan-2-one: CH₃-CO-CH₂-CH(CH₃)₂. No chiral center.
• 3-methylpentan-3-one: (CH₃CH₂)₂C(O)CH₃? Actually, it is CH₃CH₂-C(CH₃)(O)-CH₂CH₃. No chiral center.
• 2-methylpentan-3-one: CH₃CH₂-CO-CH(CH₃)₂. No chiral center? Carbon 3 is attached to two methyls? Actually, it is achiral.
• 3,3-dimethylbutan-2-one: (CH₃)₃C-CO-CH₃. No chiral center.

So, only 3-methylpentan-2-one has stereoisomers (enantiomers). So total isomers: 7 constitutional, but one has 2 stereoisomers, so total 8 isomers.

Step 4: Reduction with NaBH₄ and racemination.
NaBH₄ reduces C=O to CH-OH. Racemic product is formed if:

• The ketone is achiral (so that the reduction is not stereoselective), and
• The reduction creates a new chiral center.

So, we need to check for each isomer:

1. Hexan-2-one: Reduction gives hexan-2-ol. The carbon 2 becomes chiral? Yes, so racemic mixture is formed. So, this gives racemic product.
2. Hexan-3-one: Reduction gives hexan-3-ol. Carbon 3 is not chiral? Because it has two ethyl groups? Actually, carbon 3 is CH(OH) with CH₃CH₂- and CH₃CH₂-, so it is achiral (identical groups). So, no chiral center formed. So, not racemic.
3. 3-methylpentan-2-one: This ketone is chiral (exists as R and S). Reduction gives 3-methylpentan-2-ol. The carbon 2 becomes chiral? Actually, carbon 2 is already chiral? Wait: In the ketone, carbon 2 is sp2 (carbonyl), so not chiral. Reduction makes it sp3. So, now there are two chiral centers: carbon 2 and carbon 3. So, the product is a pair of diastereomers? Actually, careful: The starting ketone is chiral (due to carbon 3). Reduction at carbon 2 (which is prochiral) will give diastereomers because the existing chiral center influences the reaction. So, not racemic mixture (diastereomers are not racemic). So, no racemic product.
4. 4-methylpentan-2-one: Reduction gives 4-methylpentan-2-ol. Carbon 2 becomes chiral? Yes. And the ketone is achiral. So, racemic product.
5. 3-methylpentan-3-one: Reduction gives 3-methylpentan-3-ol. Carbon 3 is tertiary alcohol? Actually, it is (CH₃CH₂)₂C(OH)CH₃. So, carbon 3 has three alkyl groups? It is not chiral. So, no new chiral center. Not racemic.
6. 2-methylpentan-3-one: Reduction gives 2-methylpentan-3-ol. Carbon 3 becomes chiral? Yes. And ketone is achiral. So, racemic product.
7. 3,3-dimethylbutan-2-one: Reduction gives 3,3-dimethylbutan-2-ol. Carbon 2 becomes chiral? Yes. Ketone is achiral. So, racemic product.

So, the isomers that give racemic products are: 1, 4, 6, 7. That is 4 isomers.

But wait: We also have the stereoisomers of 3-methylpentan-2-one? They are reacted separately. But as we saw, they give diastereomers, not racemates. So, they are not counted.

So, total number of ketones that give racemic product is 4.

Final Answer: 4

Related Topics and Formulae

Racemic Mixture: A 1:1 mixture of two enantiomers. Formed when an achiral reactant undergoes a reaction that creates a chiral center without stereochemical bias.

Reduction of Ketones: NaBH₄ reduces ketones to secondary alcohols. The reaction is not stereoselective for achiral ketones, leading to racemic mixtures if a new chiral center is formed.

Molecular Formula Calculation: For ketone C_nH_2nO, MW = 14n + 16.