Dehydrohalogenation of 1-chloro-1-methylcyclopropane affords two alkenes (A and B) as products.

Chemistry

1 answer:

UNO [17]3 years ago

4 0

Explanation:

Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:

Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.

However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.

Elimination of the hydrogen from the methyl group is easier.

Thus, the major product will A

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What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n

ANEK [815]

Answer: The volume of permanganate ion (potassium permanganate) is 10.0 mL

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol$

For the given chemical equation:

$5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O$

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = $\frac{1}{5}\times 0.00762=0.00152mol$ of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

$0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL$