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Effectus [21]
2 years ago
6

Arrange the following alkyl bromides in order from most reactive to least reactive in an SN2 reaction: 1-bromo-2-methylbutane, 1

-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.
Rank the alkyl bromides from most reactive to least reactive. To rank items as equivalent, overlap them.
And what makes an alkyl bromide more reactive in this case?

Chemistry
2 answers:
xxMikexx [17]2 years ago
5 0
Primary alkyl halides tend to undergo the SN2 reaction mechanism in nucleophilic substitution since there is less steric hindrance for nucleophilic attack and the carbocations that they form are not as stable as those formed from tertiary alkyl halides. 


1-bromopentane > 1-bromo 2-methylbutane > <span>1-bromo-3-methylbutane</span>> 2-bromo 2-methylbutane
Law Incorporation [45]2 years ago
5 0

Answer:

See explanation

Explanation:

First, in order to know this it's neccesary to remember how a SN2 reaction takes place. A Sn2 reaction is a bimolecular concerted reaction where all bonds are broken and making in only one step.

For this to occur, we need a strong nucleophyle (such a strong base) and a substract with a great outgoing group (The halides are great leaving groups).

The nucleophyle attacks on the back side of the molecule with the bromine, and the result is a molecule with inverted configuration and the bromine is replaced by the nucleophyle.

However, this step is fast and concerted, and in order to do this faster, the reactant must be (Ideally) with no substituent, because if the molecule is bulky, the nucleophyle's attack to the back side is hard. This doesn't mean that it will not undergo, but it will be harder and slower.

Because of this reason, we can see that from all the alkyl bromides there, the least bulky is the 1-bromopentane, so this will be the more reactive in Sn2, followed by 1-bromo-methylbutane, then the 1-bromo-2-methylbutane and finally the 2 - bromo - 2 - methylpentane.

In the picture, you have the structures of these molecules, so you can see how the steric hindrance affects this.

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6 0
3 years ago
A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu
vichka [17]

Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 560.00 mmHg

P_2 = final pressure = ?

V_1 = initial volume = 280 mL

V_2 = final volume = 560.0 mL

Now put all the given values in the above formula, we get:

560.00mmHg\times 280 mL=P_2\times 560.0 mL

P_2=280mmHg

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

3 0
3 years ago
The students in the picture below are using a globe and a lamp to model the Sun and the Earth. If the model Earth acts the same
timofeeve [1]

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The only one that makes sense IF the model behaves as the Earth is D.

Explanation:

7 0
3 years ago
The compound ncl3 is nitrogen trichloride, but alcl3 is simply aluminum chloride. why
AveGali [126]
The answer to this question would be:
NCl3 is a molecular compound (two or more nonmetals), and therefore in its name prefixes indicate the number of each type of atom. so NCl3 is nitrogen trichloride<span>.
</span><span>The compound AlCl3 is an ionic compound (metal and nonmetal), and therefore does not require prefixes. so AlCl3 is aluminum chloride.
</span><span>
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5 0
3 years ago
Read 2 more answers
A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

Best regards.

6 0
3 years ago
Read 2 more answers
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