Give the decreasing order of hyperconjugative effect of `R` in `R - CH = CH_2`, where `R` is :
(I) `Me-`
(II) `Et-`
(III) `Me_2CH-`
(IV) `Me_3C-`.

Options

A. `I gt II gt III gt IV`
B. `IV gt III gt II gt I`
C. `II gt I gt III gt IV`
D. `IV gt III gt I gt II`

Explanation

To determine the decreasing order of hyperconjugative effect of `R` in the compound `R - CH = CH_2`, we need to analyze how many alpha hydrogens are present in each case. The hyperconjugative effect is directly related to the number of alpha hydrogens; more alpha hydrogens lead to a greater hyperconjugative effect.

Let's evaluate each substituent `R` one by one:

1. **For R = Me- (Methyl group, CH₃-)**:
- The structure becomes: `CH₃ - CH = CH₂`
- The alpha carbon (the carbon directly attached to the double bond) has 3 alpha hydrogens (from the methyl group).
- **Alpha Hydrogens = 3**

2. **For R = Et- (Ethyl group, CH₃CH₂-)**:
- The structure becomes: `CH₃CH₂ - CH = CH₂`
- The alpha carbon has 2 alpha hydrogens (from the ethyl group).
- **Alpha Hydrogens = 2**

3. **For R = Me₂CH- (Isopropyl group, (CH₃)₂CH-)**:
- The structure becomes: `CH₃CH(CH₃) - CH = CH₂`
- The alpha carbon has 1 alpha hydrogen (the one on the isopropyl group).
- **Alpha Hydrogens = 1**

4. **For R = Me₃C- (Tert-butyl group, (CH₃)₃C-)**:
- The structure becomes: `CH₃C(CH₃)₂ - CH = CH₂`
- The alpha carbon has 0 alpha hydrogens (the tert-butyl group has no hydrogens attached to the alpha carbon).
- **Alpha Hydrogens = 0**

Now, we can summarize the number of alpha hydrogens for each substituent:
- Me-: 3 alpha hydrogens
- Et-: 2 alpha hydrogens
- Me₂CH-: 1 alpha hydrogen
- Me₃C-: 0 alpha hydrogens

### Conclusion:
The decreasing order of hyperconjugative effect based on the number of alpha hydrogens is:
1. Me- (3 alpha hydrogens)
2. Et- (2 alpha hydrogens)
3. Me₂CH- (1 alpha hydrogen)
4. Me₃C- (0 alpha hydrogens)

Thus, the final order is:
**Me- > Et- > Me₂CH- > Me₃C-**