Each of the organic compounds mentioned has a general formula so that we can identify the classification of a certain substance. The compound CH₃CH₂OH is an alcohol because it follows the general formula R-OH, where R is a hydrocarbon chain. In this case, the hydrocarbon chain is ethane. When a hydroxyl functional group is attached, it becomes an alcohol whose name is ethanol.
Explanation:
brain and heart froce acts between two chloramine
From the combustion of octane, the formaldehyde will be formed as this equation:
C8H18 + O2 → CH2O + H2O this is the original equation but it is not a balanced equation, so let's start to balance it:
the equation to be balanced so the number of atoms on the right side of the equation sholud be equal with the number of atoms on the lef side.
-we have 8 C atoms on left side and 1 atom on the right side so we will try putting 8 CH2O on the right side instead of CH2O
C8H18 + O2 → 8 CH2O + H2O
we have 2 O atoms on the left side and 9 atoms on the right side so we will try first to put 9 O2 instead of O2 on the left side and put 2H2O on the right side and put 16 CH2O instead of 8 CH2O to make the atoms of O are equal on both sides = 18 atoms
C8H18 + 9 O2 → 16 CH2O + 2H2O
put now we have 8 atom C on the left side and 16 atom on the right side so, we will put 2 C8H18 instead of C8H18 now we get this equation:
2C8H18 + 9O2 →16 CH2O + 2H2O
-now we have 36 of H atoms on both sides.
- and 16 of C atoms on both sides.
- and 18 of O atoms on both sides.
now all the number of atoms of O & C & H are equal on both sides
∴ 2C8H18 + 9O2 → 16 CH2O + 2 H2O
is the final balanced equation for the formation of formaldehayde
Answer:
See explanation
Explanation:
The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.
On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.
