Consider all the structural isomers with molecular formula $$C_{5}H_{11}Br$$ are separately treated with KOH(aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is ______.

We need to find all structural isomers of $$C_5H_{11}Br$$, treat each with aqueous KOH (SN2/SN1 substitution, no rearrangement), and count how many products exhibit optical isomerism.

The structural isomers of $$C_5H_{11}Br$$ (monobromopentanes) are:

1. $$CH_3CH_2CH_2CH_2CH_2Br$$ (1-bromopentane)

2. $$CH_3CH_2CH_2CHBrCH_3$$ (2-bromopentane)

3. $$CH_3CH_2CHBrCH_2CH_3$$ (3-bromopentane)

4. $$(CH_3)_2CHCH_2CH_2Br$$ (1-bromo-3-methylbutane)

5. $$(CH_3)_2CHCHBrCH_3$$ (2-bromo-3-methylbutane)

6. $$(CH_3)_2CBrCH_2CH_3$$ (2-bromo-2-methylbutane)

7. $$(CH_3)_3CCH_2Br$$ (1-bromo-2,2-dimethylpropane / neopentyl bromide)

8. $$CH_3CH_2CH(CH_3)CH_2Br$$ (1-bromo-2-methylbutane)

Now, substitution with KOH(aq) replaces Br with OH. The product is the corresponding alcohol. For optical isomerism, the product must have a chiral center (asymmetric carbon).

1. 1-bromopentane → 1-pentanol: $$CH_3CH_2CH_2CH_2CH_2OH$$ — No chiral center. Not optically active.

2. 2-bromopentane → 2-pentanol: $$CH_3CH(OH)CH_2CH_2CH_3$$ — C2 has four different groups (H, OH, CH3, CH2CH2CH3). Chiral. Optically active.

3. 3-bromopentane → 3-pentanol: $$CH_3CH_2CH(OH)CH_2CH_3$$ — C3 has groups H, OH, CH2CH3, CH2CH3. Two groups are the same. Not chiral. Not optically active.

4. 1-bromo-3-methylbutane → 3-methyl-1-butanol: $$(CH_3)_2CHCH_2CH_2OH$$ — No chiral center. Not optically active.

5. 2-bromo-3-methylbutane → 3-methyl-2-butanol: $$(CH_3)_2CHCH(OH)CH_3$$ — C2 has groups H, OH, CH3, CH(CH3)2. All four different. Chiral. Optically active.

6. 2-bromo-2-methylbutane → 2-methyl-2-butanol: $$(CH_3)_2C(OH)CH_2CH_3$$ — C2 has groups OH, CH3, CH3, CH2CH3. Two CH3 groups are same. Not chiral. Not optically active.

7. 1-bromo-2,2-dimethylpropane → 2,2-dimethyl-1-propanol (neopentyl alcohol): $$(CH_3)_3CCH_2OH$$ — No chiral center. Not optically active.

8. 1-bromo-2-methylbutane → 2-methyl-1-butanol: $$CH_3CH_2CH(CH_3)CH_2OH$$ — C2 (the carbon bearing CH3) has groups H, CH3, CH2CH3, CH2OH. All four different. Chiral. Optically active.

The products that can exhibit optical isomerism are: 2-pentanol (from isomer 2), 3-methyl-2-butanol (from isomer 5), and 2-methyl-1-butanol (from isomer 8).

The answer is 3.