What happens if two deactivating and one activating group is present on a benzene ring?

We know that reactivity of benzene ring towards electrophilic reaction depends upon presence of group on benzene ring. Benzene gives electrophilic substitution reaction due to negative pi cloud formation , if there is EWG (electron withdrawing) group on benzene ring it's negativeness decreases and reactivity also decreases and due to resonance upcoming group goes at meta position only while in the presence of ERG (electron releasing groups) it's negative ness increases and reactivity increase and due to resonance upcoming group goes at ortho and para position .

If both EWG and ERG present on benzene ring then upcoming electrophile goes to ortho and para position  according to ERG group .

What is bromoform reaction?

Haloform reaction includes Chloroform, Bromoform and Iodoform reaction, It is feature reaction of terminal methyl ketones and alpha hydroxy ketones . In which methyl ketones oxidised in the presence of I2 and NaOH ( NaOI) into Sodium salt of carboxylic acid and corresponding haloform ( Chloroform , Bromoform and Iodoform)

Is Methanal an aldehyde?

Yes ! Methanal is a aldehyde containing -CHO group. And is molecular formula is HCHO . It is more commonly known as formaldehyde. It is also first member of aldehyde homologous series.

Which is a better nucleophile, among halides ions ( fluoride, Chloride, bromide and iodide)? and Why?

Nucleophilicity of an nucleophilic depends upon nature of solvent, for example:

(1) I⁻ is a better nucleophile than F⁻ in polar protic solvents.

(2) F⁻ is a better nucleophile than Br⁻ in polar aprotic solvents.

We know that a protic solvent has an H atom bound to more electronegative elements like F, O or N. It can use its H atom to formed H-bonding with a nucleophile. Which accumulate around nucleophilic and creates a shell around the nucleophile. So that it becomes more difficult to attack the positive carbon bearing the leaving group.

F⁻ is a small ion with a high charge density. It is tightly solvated.

I⁻ is a large ion with a low charge density. It is loosely solvated. There are only a few solvent molecules to push out of the way.

Hence over all order of nucleophilicity in polar protic solvents is given as :

        [ I⁻ > Br⁻ > Cl⁻ > F⁻ ]

We know that a polar aprotic solvent does not have a hydrogen atom that can formed hydrogen bond.

But in all of them, the negative ends of the dipoles directedaway from the molecule. So that It is easy for them to solvate cations.

The positive ends of the dipoles are closer to the middle of the molecule. It is difficult for them to get close to the anions.

As the result the nucleophile has few molecules in its solvent shell. The nucleophile can more easily attack on positive carbon (electrophilic center).

Hence F⁻ becomes a much better nucleophile than chloride, 

Hence over all order of nucleophilicity in polar aprotic solvents is just reverse of that given in polar protic solvent :

        [ F⁻ > Cl⁻ > Br⁻ > I⁻ ]

Why is BrO4- a stronger oxidising agent than ClO4-?