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3 years ago

An alkane with the formula C6H14 can be prepared by hydrogenation of either of two precursor alkenes having the formula C6H12.

Chemistry

2 answers:

PIT_PIT [208]3 years ago

7 0

Answer:

2-methylpentane

Explanation:

Hello,

On the attached document, you will find the solution for this exercise. It is important for you to notice that at the reactions there are two possible precursors, 2-methylpentene and 4-methylpentene which are suitable to form 2-methylpentane via hydrogenation.

Best regards.

ankoles [38]3 years ago

6 0

Two precursor alkenes

H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene

H₃C   CH₃
I   I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene

alkane

H₃C CH₃
  I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane

H₃C  CH₃ H₃C CH₃
I   I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

H₃C CH₃ H₃C   CH₃
I    I     I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

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Explanation:

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enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

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