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Answer 1 · 2018-02-16T13:13:00

Alkenyl radicals with formula C4H7 are seven

Consider first the alkenes with formula C4H8:
They are
1) H2C=CH-CH2-CH3
2) H3C-CH=CH-CH3
3) H2C=C(CH3)2

From 1 there are three radicals possibles:

1a) H2C=CH-CH2-CH2
1b) H2C=CH-CH-CH3
1c) H2C=C*-CH2-CH3

From 2 the possibilities are:

2a) H3C-CH=CH-CH2
2b) H3C-CH=C-CH3

From 3:

3a) $H_{3} {C-C=CH}_{2}$ $"H"_3"C-C=CH"_2$
$m m m m l l ∣$ $"color(white)(mmmmll)|$
$m m m m l {CH}_{2} *$ $color(white)(mmmml)"CH"_2"*"$

3b) $H_{3} C-C=CH*$ $"H"_3"C-C=CH*"$
$m m m m l l ∣$ $color(white)(mmmmll)|$
$m m m m l {CH}_{3}$ $color(white)(mmmml)"CH"_3$

In total seven radicals

Answer 2 · 2018-02-16T16:01:37

I count ten alkenyl radicals with the formula $C_{4} H_{7}$ $"C"_4"H"_7$ .

From but-1-ene

1. $"H"stackrelcolor(blue)(bb·)("C")"=CHCH"_2"CH"_3$

2. $"H"_2"C="stackrelcolor(blue)(bb·)("C")"CH"_2"CH"_3$

3. $"H"_2"C=CH"stackrelcolor(blue)(bb·)("C")"HCH"_3$

4. $"H"_2"C=CHCH"_2stackrelcolor(blue)(bb·)("C")"H"_2$

From but-2-ene

5. and 6. $stackrelcolor(blue)(bb·)("C")"H"_2"CH=CHCH"_3$ ( $E"-"$ and $Z"-"$ isomers)

7. and 8. $"CH"_3stackrelcolor(blue)(bb·)("C")"=CHCH"_3$ ( $E"-"$ and $Z"-"$ isomers)

From 2-methylpropene

9. $stackrelcolor(blue)(bb·)("C")"H"_2"C"("CH"_3)"=CH"_2$

10. $("CH"_3)_2"C="stackrelcolor(blue)(bb·)("C")"H"$