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Elina [12.6K]
3 years ago
10

During the bromination of methane, the free radical CH3• is generated. A possible terminating step of this reaction is the forma

tion of C2H6 by the combination of two free radicals.
What could be produced in a terminating step during the bromination of propane ?

Chemistry
2 answers:
podryga [215]3 years ago
7 0

Answer:

There will be formation of 2,3-dimethylbutane.

Explanation:

Free radical reactions are more random reactions as compared to ionic reactions.

When a propane molecules undergoes bromination there may be removal of primary hydrogen or secondary hydrogen (as shown in figure). The removal of primary hydrogen generates primary carbon free radical which is less stable than secondary carbon free radical (this can be attributed to more hyperconjugative structures in secondary carbon free radical).

So the more stable free radical will be the dominate or abundant free radical during the reaction. This will combine with another secondary free radical during termination step(as shown in figure).

aalyn [17]3 years ago
3 0

The answer is (CH₃)₂CH-CH(CH₃)₂.

That is the free radical generated in bromination of propane is (CH₃)₂CH° and CH₃CH₂CH₂° , but due to more stability of secondary free radical , (CH₃)₂CH° is more stable than that of CH₃CH₂CH₂°, so reaction proceeds via (CH₃)₂CH° radical. and than at termination the the product formed will be (CH₃)₂CH-CH(CH₃)₂.

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<u>d</u><u>.</u><u> </u><u>amine</u><u>.</u>

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neonofarm [45]

Answer:

\large \boxed{7.62}

Explanation:

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\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 7.54 +\log \left(\dfrac{0.450}{0.450}\right )\\\\& = & 7.54 + \log1.00 \\ & = & 7.54 + 0.00\\& = & 7.54\\\end{array}

2. pH after adding strong base

(a) Find new composition of the buffer

The base reacts with the HA and forms A⁻

\text{ Initial moles of HA} = \text{1.0 L} \times \dfrac{\text{0.450 mol }}{\text{1.00 L}} = \text{0.450 mol = 450 mmol}\\\\\text{ Initial moles of A}^{-} = \text{1.0 L} \times \dfrac{\text{0.450 mol }}{\text{1.00 L}} = \text{0.450 mol = 450 mmol}\\\\\text{Moles of OH$^{-}$ added} = \text{40 mmol}

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I/mmol:    450            40              450

C/mmol:    -40           -40                 40

E/mmol:    410               0              490

(b) Find the new pH  

\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 7.54 +\log \left(\dfrac{490}{410}\right )\\\\& = & 7.54 + \log1.195 \\& = & 7.54 +0.0774\\& = & \mathbf{7.62}\\\end{array}\\\text{The new pH is $\large \boxed{\textbf{7.62}}$}

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PH = - log [ H+ ]

pH = - log [ 8.2 x 10⁻⁴]

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Answer:

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