The increasing order of basicity for the following intermediates is (from weak to strong)
`H_(3)C-underset("(i)")underset(CH_(3))underset("| ")overset(CH_(3))overset("| ")(C^(Theta))" "underset("(ii)")(H_(2)C=CH-overset(Theta)CH_(2))" "underset("(iii)")(HC=overset(Theta)C)" "underset("(iv)")(overset(Theta)CH_(3))" "underset("(v)")(overset(Theta)CN)`

Options

A. `(v) lt (iii) lt (ii) lt (iv) lt(i)`
B. `(iii) lt (iv) lt (ii) lt (i) lt (v)`
C. `(v) lt (i) lt (iv) lt (ii) lt (iii)`
D. `(iii) lt (i) lt (ii) lt (iv) lt (v)`

Answer

To determine the increasing order of basicity for the given intermediates, we need to analyze the stability of the carbanions formed in each case. Basicity is inversely related to the stability of the carbanion: the more stable the carbanion, the less basic it is.

### Step-by-Step Solution:

1. **Identify the Structures**:
- (i) \( \text{H}_3C- \overset{(i)}{C} \overset{(i)}{(CH_3)} \)
- (ii) \( \text{H}_2C=CH-\overset{(ii)}{C} \overset{(ii)}{(H_2)} \)
- (iii) \( \text{HC} \equiv \overset{(iii)}{C} \)
- (iv) \( \overset{(iv)}{C} \text{H}_3 \)
- (v) \( \overset{(v)}{C} \text{N} \)

2. **Analyze Hybridization**:
- The hybridization of the carbon atoms in each structure affects the stability of the carbanion.
- **(i)**: The carbon is \( sp^3 \) hybridized.
- **(ii)**: The carbon is \( sp^2 \) hybridized.
- **(iii)**: The carbon is \( sp \) hybridized.
- **(iv)**: The carbon is \( sp^3 \) hybridized.
- **(v)**: The carbon is \( sp \) hybridized.

3. **Stability of Carbanions**:
- Carbanions with \( sp \) hybridization are more stable due to the higher electronegativity of the \( sp \) hybridized carbon, making the negative charge more stable.
- Carbanions with \( sp^2 \) hybridization are less stable than \( sp \) but more stable than \( sp^3 \).
- Carbanions with \( sp^3 \) hybridization are the least stable.

4. **Order of Basicity**:
- **(v)**: \( sp \) hybridized carbanion is the most stable, hence the least basic.
- **(iii)**: \( sp \) hybridized carbanion is also stable but less than (v).
- **(ii)**: \( sp^2 \) hybridized carbanion is less stable than (iii) and (v), hence more basic.
- **(iv)**: \( sp^3 \) hybridized carbanion is less stable than (ii), hence more basic.
- **(i)**: \( sp^3 \) hybridized carbanion is the least stable, hence the most basic.

5. **Final Order**:
- From weak to strong basicity: **(v) < (iii) < (ii) < (iv) < (i)**

### Conclusion:
The increasing order of basicity for the given intermediates is:
**(v) < (iii) < (ii) < (iv) < (i)**