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sertanlavr [38]
3 years ago
8

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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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
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Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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3 years ago
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