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Oksi-84 [34.3K]
3 years ago
12

Draw the structure of two alkenes that would yield 1-methylcyclohexanol when treated with Hg(OAc)2 in water, then NaBH4:

Chemistry
2 answers:
ratelena [41]3 years ago
7 0

The possible alkenes to produce 1-methylcyclohexanol are  and . (Refer to the attached image)

Further explanation:

Oxymercuration-demercuration reaction:

Oxymercuration-demercuration converts alkene into alcohol in two steps. In the initial step, the alkene is reacted with {\text{H}}{{\text{g}}^{2 + }} salt and nucleophile such as alcohol and water to produce an organomercury intermediate. In second step, the carbon-Hg bond in organomercury intermediate is converted into C-H bond.

In this method the product is obtained in accordance with Markovnikov’s rule, and rearrangement of carbocation does not take place.

The mechanism of Oxymercuration-demercuration is as follows(Refer to the attached image):

Step 1: {\text{Hg}}{\left( {{\text{OAc}}} \right)_2} break and form mercuroacetic cation \left( {{\text{HgOA}}{{\text{c}}^ + }} \right) and {\text{OA}}{{\text{c}}^ - } ions.

Step 2: The mercuroacetic cation reacts with a double bond and forms a cyclic cation.

Step 3:The cyclic intermediate is attacked by the {{\text{H}}_{\text{2}}}{\text{O}} nucleophile to the most substituted side according to Markovnikov’s rule and forms organomercury intermediate.

Step 4: the carbon-Hg bond in organomercury intermediate is converted into C-H bond.

The methylcyclohexanol has alcohol on tertiary carbon thus the double bond must be attached to the tertiary carbon in initial alkene. This means there are only two possible alkenes that can have double bond attached to the tertiary carbon. These alkenes are 1-methylcyclohexene and methylenecyclohexane.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603

2. Determine the product formed when 2-propanol react with NaH: brainly.com/question/5045356

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter:Oxymercuration-demercuration reactions

Keywords: Oxymercuration-demercuration reaction, alkenes, structure of two alkenes, yield 1-methylcyclohexanol, Hg(OAc)2, NaBH4.

babymother [125]3 years ago
5 0

Answer:

a) methylidenecyclohexane

b) 1-methylcyclohex-1-ene

Explanation:

The given reaction is oxymercuration-demercuration.

This reaction gives alcohol (according to Markovnikov's rule) from alkenes.

The reaction of alkene to give desired  1-methylcyclohexanol is shown in the figure.

The alkenes are:

a) methylidenecyclohexane

b) 1-methylcyclohex-1-ene

Both will give alcohol where hydroxyl group will be on the more substituted double bonded carbon atom.

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Explanation:

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= Atomic Weight of hydrogen + Atomic Weight  of Bromine

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What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac
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To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

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• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

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3 years ago
Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium
pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

k_{c}=3.07\times10^{-4}

Temperature = 24°C

We know that,

The balance equation is

2NOBr\Rightarrow 2NO+Br_{2}

Initial concentration is,

0.878\Rightarrow 0+0

Concentration is,

-2x\Rightarrow 2x+x

Equilibrium concentration

0.878-2x\Rightarrow 2x+x

We need to calculate the value of x

Using formula of concentration

k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}

Put the value into the formula

3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}

2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}

2x^2=0.0002367+0.001228x^2-0.0010536x

2x^2-0.001228x^2+0.0010536x-0.0002367=0

1.998772x^2+0.0010536x-0.0002367=0

x=0, 0.01062

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

concentration\ of\ NO=2x

Put the value of x

concentration\ of\ NO=2\times0.01062

concentration\ of\ NO=0.02124

Hence, The equilibrium concentration of NO is 0.02124 M.

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