Consider the following reactions
(1) `(CH_3)_3"CC"H(OH)CH_3overset("conc".H_2SO_4)rarr`
(2) `(CH_3)_2"CHCH(Br)CH_(3)overset(conc.KOH)rarr`
(3) `(CH_3)_2"CHCH(Br)CH_(3)overset((CH_3)_3CO^(-) Koplus)rarr`
(4) `(CH_3)_2underset( "OH ")underset(" | ")(C-CH_2)-CHOoverset(Delta)rarr`
Which of these reaction(s) will produce Saytzeff product?

Options

A. (1) , (2) and (4)
B. (3) only
C. (4) only
D. (2) and (4)

Answer

To determine which of the given reactions produce the Saytzeff product, we will analyze each reaction step by step.

### Reaction 1:
**Reaction:** \((CH_3)_3C(OH)CH_3 \overset{\text{conc. } H_2SO_4}{\rightarrow}\)

1. **Protonation of Alcohol:** The hydroxyl group (OH) is protonated by concentrated sulfuric acid, forming a better leaving group (water).
2. **Formation of Carbocation:** The water molecule leaves, leading to the formation of a tertiary carbocation \((CH_3)_3C^+\).
3. **Carbocation Rearrangement:** Since we have a tertiary carbocation, it is stable and does not rearrange further.
4. **Deprotonation:** A proton is removed from a neighboring carbon, resulting in the formation of a double bond. The double bond forms between the carbon with the carbocation and the adjacent carbon.
5. **Product:** The product is a double bond between the carbons, leading to a more substituted alkene (Saytzeff product).

### Reaction 2:
**Reaction:** \((CH_3)_2CHCH(Br)CH_3 \overset{\text{conc. KOH}}{\rightarrow}\)

1. **Base Abstraction:** The concentrated KOH acts as a strong base and abstracts a proton from the carbon adjacent to the carbon bonded to bromine (Br).
2. **Elimination Reaction:** The Br leaves as a leaving group, resulting in the formation of a double bond.
3. **Product:** The product is a double bond between the carbons, leading to a more substituted alkene (Saytzeff product).

### Reaction 3:
**Reaction:** \((CH_3)_2CHCH(Br)CH_3 \overset{(CH_3)_3CO^-}{\rightarrow}\)

1. **Bulky Base Action:** The bulky base \((CH_3)_3CO^-\) will abstract a proton from the less substituted carbon due to steric hindrance.
2. **Elimination Reaction:** The Br leaves, and a double bond is formed.
3. **Product:** This results in a less substituted alkene (Hofmann product), not a Saytzeff product.

### Reaction 4:
**Reaction:** \((CH_3)_2CHO \overset{(C-CH_2)-CHO \Delta}{\rightarrow}\)

1. **Heating with Aldehyde:** The reaction involves heating with an aldehyde, which can lead to dehydration.
2. **Formation of Double Bond:** The reaction can lead to the formation of a double bond.
3. **Product:** Depending on the structure, this can lead to a more substituted alkene (Saytzeff product).

### Conclusion:
From the analysis:
- **Reactions 1, 2, and 4** produce Saytzeff products.
- **Reaction 3** produces a Hofmann product.

### Final Answer:
The reactions that produce Saytzeff products are **1, 2, and 4**.