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3 years ago

Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.

Chemistry

1 answer:

Nimfa-mama [501]3 years ago

5 0

<h2>Heptene formed is -</h2><h2> $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$ </h2>

Explanation:

The two possibilities when the peroxide is not present

$CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C$ ≡ $CH +HBr$ → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}$

$CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3$

In presence peroxide,

$CH_3-CH_2-CH_2-CH_2-CH_2-C$ ≡ $CH$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.

This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
One mole of HBr adds to one mole of 1-heptane.
The structure of heptene formed is -

$CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

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the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of B

ddd [48]

Answer:

1.8 × 10⁻⁴ mol M/s

Explanation:

Step 1: Write the balanced reaction

2 Br⁻ ⇒ Br₂

Step 2: Establish the appropriate molar ratio

The molar ratio of Br⁻ to Br₂ is 2:1.

Step 3: Calculate the rate of appearance of Br₂

The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:

3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s

3 0

3 years ago

A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$

So we solve for C2:

$C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M$

2. Now, since 111 mL of water is added, we compute the final volume, V3:

$V_3=139+111=250mL$

So, the final concentration of the 139 mL portion is:

$C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M$

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3 years ago

Calculate the equilibrium constant of the reaction below if the pressures are 1.0atm, 2.0 atm, and 1.0 atm respectively. PCl3 +

Makovka662 [10]

Answer:

K = 0.5

Explanation:

Based on the reaction:

PCl₃ + Cl₂ ⇄ PCl₅

The equilibrium constant, K, is defined as:

K = P PCl₅ / P PCl₃ * P Cl₂

<em>Where P represent the pressure at the equilibrium for each one of the gases involved in the equilibrium.</em>

<em />

As:

P PCl₅ = 1.0atm

P PCl₃ = 1.0atm

P Cl₂ = 2.0atm

K = 1.0atm / 1.0atm * 2.0atm

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2 years ago

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

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The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

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